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16 Sep 2014, 00:37
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A
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Is the average (arithmetic mean) of data set S greater than its median?
(1) Data set S consist of consecutive multiples of 3
(2) The sum of all terms of data set S is 75
(1) Data set S consist of consecutive multiples of 3
(2) The sum of all terms of data set S is 75
16 Sep 2014, 00:37
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Official Solution:
Is the average (arithmetic mean) of data set S greater than its median?
(1) Data set S consist of consecutive multiples of 3.
Data set S is evenly spaced. One of the most important properties of an evenly spaced set (also known as an arithmetic progression) is: in any evenly spaced set, the arithmetic mean (average) is equal to the median. Therefore, the mean of \(S\) = the median of \(S\). This statement is sufficient.
(2) The sum of all terms of data set S is 75.
If \(S=\{75\}\), then the mean = median. However, if \(S=\{-1, 1, 75\}\), then the mean (25) is greater than the median (1). This statement is not sufficient.
Answer: A
Is the average (arithmetic mean) of data set S greater than its median?
(1) Data set S consist of consecutive multiples of 3.
Data set S is evenly spaced. One of the most important properties of an evenly spaced set (also known as an arithmetic progression) is: in any evenly spaced set, the arithmetic mean (average) is equal to the median. Therefore, the mean of \(S\) = the median of \(S\). This statement is sufficient.
(2) The sum of all terms of data set S is 75.
If \(S=\{75\}\), then the mean = median. However, if \(S=\{-1, 1, 75\}\), then the mean (25) is greater than the median (1). This statement is not sufficient.
Answer: A
General Discussion
04 Feb 2016, 08:50
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Hello Friend,
The question asks "Is the mean of set greater than its median?"
In a set of even numbers (Example: 3, 6, 9, 12). The mean (avg) is 7.5.... The median, stated by your example, will also be 7.5 (mean of the two middle numbers 6, 9).
Thus the mean equals the median (you state this as well). It is not greater than the median as the original question states. I would deem this: not sufficient.
If the original questions asked if the mean of set is greater or equal to its median, then it would be sufficient.
Thoughts?
The question asks "Is the mean of set greater than its median?"
In a set of even numbers (Example: 3, 6, 9, 12). The mean (avg) is 7.5.... The median, stated by your example, will also be 7.5 (mean of the two middle numbers 6, 9).
Thus the mean equals the median (you state this as well). It is not greater than the median as the original question states. I would deem this: not sufficient.
If the original questions asked if the mean of set is greater or equal to its median, then it would be sufficient.
Thoughts?
